Find a basis for `QQ(root (3)(2))` over `QQ`.

Reformulate the Given Exercise: The task is to find a set of elements B that forms a basis for the field extension `QQ(root (3)(2))` when it is viewed as a vector space over the field `QQ` (the rational numbers). We must find a set B and , implicitly , the dimension `[QQ(root (3)(2)) : QQ ]`.

General Strategy for a Solution:

1. Identify the Element: Let `alpha = root (3)(2)`.
2. Find the Minimal Polynomial: Determine the non-zero polynomial p(x) with coefficients in `QQ` of the lowest degree such that `alpha` is a root.
3. Determine the Dimension: The degree of the minimal polynomial of `alpha` over `QQ` gives the dimension of the vector space , `n = [QQ(alpha) : QQ]`.
4. Construct the basis: The basis B will be the set of powers of `alpha` up to n - 1 : `B = { 1 , alpha , alpha^2 , ... , alpha^{n-1}}`.

Specific Theory Used:

Field Extension `QQ(alpha)`: The smallest field containing both `QQ` and the algebraic number `bb{alpha}`.
Minimal Polynomial: The unique monic , irreducible polynomial `p(x) in QQ[x]` such that `bb{p(alpha) = 0}`.
Dimension of a Simple Extension: The degree of the field extension , `bb{[QQ(alpha) : QQ]}` , is equal to the degree of the minimal polynomial of `alpha` over `QQ`.
Basis for Field Extension: If `[QQ(alpha) : QQ] = n` , then the basis for `QQ(alpha)` is `bb{B = {1 , alpha , alpha^2 , ... , alpha^{n-1}}}`.

Do the Exercise:

Step1 : Determine the Minimal Polynomial and Dimension:
1. The algebraic number is `bb{alpha = root(3)(2)}`.
2. We look for the simplest polynomial `bb{p(x) in QQ[x]}` such that `p(alpha) = 0 : x = root(3)(2) implies x^3 = 2 implies bb{x^3 - 2 = 0}`.
3. The polynomial `p(x) = x^2 - 2` is monic and has rational coefficients.
4. To ensure it is the minimal polynomial , we must confirm it is irreducible over `QQ`. By Eisenstein's Criterion with prime p = 2 :
- 2 divides the coefficient of `x^0` (which is -2).
- 2 divides the coefficient of `x^1` and `x^2` (which are 0).
- The leading coefficient 1 is not divisible by 2.
- `2^2 = 4` does not divide the coefficient of `x^0` (which is -2).
Therefore , `p(x) = x^3 - 2` is irreducible over `QQ`.
5. The degree of the minimal polynomial is n = 3.
6. The dimension of the vector space is `bb{[QQ(root(3)(2))] = 3}`.
Step 2: Construct the basis:
1. Since the dimension n = 3 , the basis set B must contain three elements.
2. Using the form `B = {1 , alpha , alpha^2 }` with `alpha = root(3)(2)` , we get: `B = {1 , root(3)(2) , (root (3)(2))^2}`.
3. Simplifying the terms gives the final basis.

Conclusion:
The basis for `QQ(root(3)(2))` over `QQ` is `bb{B = {1 , root(3)(2) , root(3)(4)}}`. Any element in `QQ(root(3)(2))` is uniquely written in the form `a + b * root(3)(2) + c * root(3)(4)` , where `a , b , c in QQ`. END